1. Chapter 15 Preview Objectives Combining Light Waves
Section 1 Interference
Preview
Objectives
Combining Light Waves
Demonstrating Interference
Sample Problem

2. Chapter 15
Section 1 Interference
Objectives
Describe how light waves interfere with each other to produce bright and dark fringes.
Identify the conditions required for interference to occur.
Predict the location of interference fringes using the equation for double-slit interference.

3. Chapter 15 Combining Light Waves
Section 1 Interference
Combining Light Waves
Interference takes place only between waves with the same wavelength. A light source that has a single wavelength is called monochromatic.
In constructive interference, component waves combine to form a resultant wave with the same wavelength but with an amplitude that is greater than the either of the individual component waves.
In the case of destructive interference, the resultant amplitude is less than the amplitude of the larger component wave.

4. Interference Between Transverse Waves
Chapter 15
Section 1 Interference
Interference Between Transverse Waves

5. Combining Light Waves, continued
Chapter 15
Section 1 Interference
Combining Light Waves, continued
Waves must have a constant phase difference for interference to be observed.
Coherence is the correlation between the phases of two or more waves.
Sources of light for which the phase difference is constant are said to be coherent.
Sources of light for which the phase difference is not constant are said to be incoherent.

6. Chapter 15 Combining Light Waves Section 1 Interference
Click below to watch the Visual Concept.
Visual Concept

7. Demonstrating Interference
Chapter 15
Section 1 Interference
Demonstrating Interference
Interference can be demonstrated by passing light through two narrow parallel slits.
If monochromatic light is used, the light from the two slits produces a series of bright and dark parallel bands, or fringes, on a viewing screen.

8. Conditions for Interference of Light Waves
Chapter 15
Section 1 Interference
Conditions for Interference of Light Waves

9. Demonstrating Interference, continued
Chapter 15
Section 1 Interference
Demonstrating Interference, continued
The location of interference fringes can be predicted.
The path difference is the difference in the distance traveled by two beams when they are scattered in the same direction from different points.
The path difference equals dsin.

10. Interference Arising from Two Slits
Chapter 15
Section 1 Interference
Interference Arising from Two Slits
Click below to watch the Visual Concept.
Visual Concept

11. Demonstrating Interference, continued
Chapter 15
Section 1 Interference
Demonstrating Interference, continued
The number assigned to interference fringes with respect to the central bright fringe is called the order number. The order number is represented by the symbol m.
The central bright fringe at q = 0 (m = 0) is called the zeroth-order maximum, or the central maximum.
The first maximum on either side of the central maximum (m = 1) is called the first-order maximum.

12. Demonstrating Interference, continued
Chapter 15
Section 1 Interference
Demonstrating Interference, continued
Equation for constructive interference
d sin  = ±m m = 0, 1, 2, 3, …
The path difference between two waves =
an integer multiple of the wavelength
Equation for destructive interference
d sin  = ±(m + 1/2) m = 0, 1, 2, 3, …
an odd number of half wavelength

13. Chapter 15 Sample Problem Interference
Section 1 Interference
Sample Problem
Interference
The distance between the two slits is mm. The second-order bright fringe (m = 2) is measured on a viewing screen at an angle of 2.15º from the central maximum. Determine the wavelength of the light.

14. Sample Problem, continued
Chapter 15
Section 1 Interference
Sample Problem, continued
Interference
1. Define
Given: d = 3.0  10–5 m
m = 2
 = 2.15º
Unknown:  = ?
Diagram:

15. Sample Problem, continued
Chapter 15
Section 1 Interference
Sample Problem, continued
Interference
2. Plan
Choose an equation or situation: Use the equation for constructive interference.
d sin  = m
Rearrange the equation to isolate the unknown:

16. Sample Problem, continued
Chapter 15
Section 1 Interference
Sample Problem, continued
Interference
3. Calculate
Substitute the values into the equation and solve:

17. Sample Problem, continued
Chapter 15
Section 1 Interference
Sample Problem, continued
Interference
4. Evaluate
This wavelength of light is in the visible spectrum. The wavelength corresponds to light of a yellow-green color.

18. Chapter 15 Preview Objectives The Bending of Light Waves
Section 2 Diffraction
Preview
Objectives
The Bending of Light Waves
Diffraction Gratings
Sample Problem
Diffraction and Instrument Resolution

19. Chapter 15
Section 2 Diffraction
Objectives
Describe how light waves bend around obstacles and produce bright and dark fringes.
Calculate the positions of fringes for a diffraction grating.
Describe how diffraction determines an optical instrument’s ability to resolve images.

20. The Bending of Light Waves
Chapter 15
Section 2 Diffraction
The Bending of Light Waves
Diffraction is a change in the direction of a wave when the wave encounters an obstacle, an opening, or an edge.
Light waves form a diffraction pattern by passing around an obstacle or bending through a slit and interfering with each other.
Wavelets (as in Huygens’ principle) in a wave front interfere with each other.

21. Destructive Interference in Single-Slit Diffraction
Chapter 15
Section 2 Diffraction
Destructive Interference in Single-Slit Diffraction

22. The Bending of Light Waves, continued
Chapter 15
Section 2 Diffraction
The Bending of Light Waves, continued
In a diffraction pattern, the central maximum is twice as wide as the secondary maxima.
Light diffracted by an obstacle also produces a pattern.

23. Chapter 15 Diffraction Gratings
Section 2 Diffraction
Diffraction Gratings
A diffraction grating uses diffraction and interference to disperse light into its component colors.
The position of a maximum depends on the separation of the slits in the grating, d, the order of the maximum m,, and the wavelength of the light, .
d sin  = ±m m = 0, 1, 2, 3, …

24. Constructive Interference by a Diffraction Grating
Chapter 15
Section 2 Diffraction
Constructive Interference by a Diffraction Grating

25. Chapter 15 Sample Problem Diffraction Gratings
Section 2 Diffraction
Sample Problem
Diffraction Gratings
Monochromatic light from a helium-neon laser ( = nm) shines at a right angle to the surface of a diffraction grating that contains lines/m. Find the angles at which one would observe the first-order and second-order maxima.

26. Sample Problem, continued
Chapter 15
Section 2 Diffraction
Sample Problem, continued
Diffraction Gratings
Define
Given:  = nm =  10–7 m
m = 1 and 2
Unknown: 1 = ? 2 = ?

27. Sample Problem, continued
Chapter 15
Section 2 Diffraction
Sample Problem, continued
Diffraction Gratings
Define, continued
Diagram:

28. Sample Problem, continued
Chapter 15
Section 2 Diffraction
Sample Problem, continued
Diffraction Gratings
2. Plan
Choose an equation or situation: Use the equation for a diffraction grating.
d sin  = ±m
Rearrange the equation to isolate the unknown:

29. Sample Problem, continued
Chapter 15
Section 2 Diffraction
Sample Problem, continued
Diffraction Gratings
3. Calculate
Substitute the values into the equation and solve:
For the first-order maximum, m = 1:

30. Sample Problem, continued
Chapter 15
Section 2 Diffraction
Sample Problem, continued
Diffraction Gratings
3. Calculate, continued
For m = 2:

31. Sample Problem, continued
Chapter 15
Section 2 Diffraction
Sample Problem, continued
Diffraction Gratings
4. Evaluate
The second-order maximum is spread slightly more than twice as far from the center as the first-order maximum. This diffraction grating does not have high dispersion, and it can produce spectral lines up to the tenth-order maxima (where sin  = ).

32. Function of a Spectrometer
Chapter 15
Section 2 Diffraction
Function of a Spectrometer
Click below to watch the Visual Concept.
Visual Concept

33. Diffraction and Instrument Resolution
Chapter 15
Section 2 Diffraction
Diffraction and Instrument Resolution
The ability of an optical system to distinguish between closely spaced objects is limited by the wave nature of light.
Resolving power is the ability of an optical instrument to form separate images of two objects that are close together.
Resolution depends on wavelength and aperture width. For a circular aperture of diameter D:

34. Resolution of Two Light Sources
Chapter 15
Section 2 Diffraction
Resolution of Two Light Sources

35. Chapter 15 Preview Objectives Lasers and Coherence
Section 3 Lasers
Preview
Objectives
Lasers and Coherence
Applications of Lasers

36. Chapter 15 Objectives Describe the properties of laser light.
Section 3 Lasers
Objectives
Describe the properties of laser light.
Explain how laser light has particular advantages in certain applications.

37. Chapter 15 Lasers and Coherence
Section 3 Lasers
Lasers and Coherence
A laser is a device that produces coherent light at a single wavelength.
The word laser is an acronym of “light amplification by stimulated emission of radiation.”
Lasers transform other forms of energy into coherent light.

38. Comparing Incoherent and Coherent Light
Chapter 15
Section 3 Lasers
Comparing Incoherent and Coherent Light
Click below to watch the Visual Concept.
Visual Concept

39. Chapter 15 Laser Section 3 Lasers
Click below to watch the Visual Concept.
Visual Concept

40. Applications of Lasers
Chapter 15
Section 3 Lasers
Applications of Lasers
Lasers are used to measure distances with great precision.
Compact disc and DVD players use lasers to read digital data on these discs.
Lasers have many applications in medicine.
Eye surgery
Tumor removal
Scar removal

41. Components of a Compact Disc Player
Chapter 15
Section 3 Lasers
Components of a Compact Disc Player

42. Incoherent and Coherent Light
Chapter 15
Section 3 Lasers
Incoherent and Coherent Light